Question #f372a

1 Answer
Apr 17, 2017

Answer:

#sf(pH=8.86)#

Explanation:

Sodium acetate is the salt of a strong base and a weak acid so we would expect it to be slightly alkali due to salt hydrolysis:

#sf(CH_3COO^(-)+H_2OrightleftharpoonsCH_3COOH+OH^-)#

For which:

#sf(K_b=([CH_3COOH][OH^-])/([CH_3COO^-]))#

Before we set up an ICE table we need to find its initial concentration.

The number of moles is given by:

#sf(n_(CH_3COONa)=m/M_r=4.0/82.034=0.0487)#

We get the initial concentration using #sf(c=n/v)#:

#sf([CH_3COONa]=0.0487/0.500=0.09752color(white)(x)"mol/l")#

Now set up the ICE table based on #sf("mol/l")#:

#sf(" "CH_3COO^(-)+H_2OrightleftharpoonsCH_3COOH+OH^-)#

#sf(I" "0.09752" "0" "0)#

#sf(C" "-x" "+x" "+x)#

#sf(E" "(0.09752-x)" "x" "x)#

#:.##sf(K_b=(x^2)/((0.09752)-x))#

Since we are given #sf(pK_a)# we use this expression:

#sf(pK_a+pK_b=14)#

#:.##sf(pK_b=14-4.74=9.26)#

This means #sf(K_b)# is a very small number, so small that we can make the approximation that #sf((0.09752-x)rArr0.09752)#

So we can write:

#sf(K_b=x^2/0.09752)#

#:.##sf(x^2=K_bxx0.09752)#

#sf(x=sqrt(K_bxx0.09752)#

Since #sf(x=[OH^-])# we can take negative logs of both sides to get:

#sf(pOH=1/2[pK_b-log(0.09752)])#

#sf(pOH=1/2[9.26-(-1.01090)])#

#sf(pOH=5.1354)#

Since #sf(pH+pOH=14)# we get:

#sf(pH=14-pOH=14-5.1354=8.86)#

As expected, the pH is slightly alkaline.